If P(h,k) be point on the parabola x = 4y², which is nearest to the point Q(0,33), then the distance

Q: Wait, why do we use the normal to find the shortest distance? Can't we just use the distance formula and minimize it?

You absolutely could use the distance formula $d^2 = (x-0)^2 + (y-33)^2$, but that leads to a much more complex optimization problem involving derivatives. The geometric property that the shortest distance from a point to a curve lies along the normal is a powerful shortcut in JEE. It transforms a calculus problem into an algebraic one, saving you precious time.

Q: I got a cubic equation $t^3 + 2t - 528 = 0$. How am I supposed to solve this during the exam without a calculator?

Do not panic! In JEE Advanced, cubic equations in such problems are almost always designed to have at least one simple integer root. Always start by testing small integers like $1, 2, -1, -2$ or factors of the constant term. In this case, testing $t=8$ quickly reveals that $512 + 16 - 528 = 0$. Trust the process and look for the simple root first.

Q: Why do we need to complete the square for the second parabola? Can't we just read the directrix from the equation?

The equation $y^2 = 4(x + y)$ is not in the standard form $Y^2 = 4AX$. The presence of the $y$ term on the right side shifts the vertex away from the origin. By completing the square, we transform it into $(y-2)^2 = 4(x+1)$, which clearly shows the vertex is at $(-1, 2)$ and the focal length $A$ is $1$. Without this, you would be misidentifying the directrix line.

Question

If $P(h,k)$ be point on the parabola $x = 4y^2$, which is nearest to the point $Q(0,33)$, then the distance of P from the directrix of the parabola $y^2 = 4(x + y)$ is equal to :

Options:

(A) 2

(B) 4

(C) 8

(D) 6

Sigma Percentile · Accepted Answer

### The Geometry of Proximity: A Journey to the Nearest Point

Welcome, future engineer. Today, we are going to unravel a beautiful problem that bridges the gap between pure geometry and algebraic elegance. We are tasked with finding the distance of a specific point $P$ on the parabola $x = 4y^2$ to the directrix of a second parabola, $y^2 = 4(x + y)$.

This is not just a calculation; it is a story of symmetry and optimization.

### Phase 1: The Hunt for the Normal

Imagine you are standing at point $Q(0, 33)$ on the Cartesian plane. You want to reach the parabola $x = 4y^2$ as quickly as possible. If you were to draw circles of increasing radius centered at $Q$, the moment the circle touches the parabola, you have found your point $P$.

Geometrically, this point of contact is unique because the radius of the circle at that moment is perpendicular to the tangent of the parabola. In other words, the shortest path is always along the normal. To find $P$, we must ensure the normal at $P$ passes through $Q$.

We start by rewriting the parabola $x = 4y^2$ as $y^2 = \frac{1}{4}x$. Comparing this to the standard form $y^2 = 4ax$, we identify $4a = \frac{1}{4}$, which gives us the focal length:

$$a = \frac{1}{16}$$

### Phase 2: The Parametric Dance

Now, we employ the power of parametric coordinates. Any point $P$ on our parabola can be expressed as $(at^2, 2at)$. Substituting our value of $a$, we get:

$$P\left(\frac{t^2}{16}, \frac{t}{8}\right)$$

The equation of the normal at any point $t$ on a parabola $y^2 = 4ax$ is given by $y + tx = 2at + at^3$. Plugging in our $a = \frac{1}{16}$, the equation becomes:

$$y + tx = \frac{t}{8} + \frac{t^3}{16}$$

This line must pass through $Q(0, 33)$. By substituting $x = 0$ and $y = 33$, the equation simplifies beautifully:

$$33 = \frac{t}{8} + \frac{t^3}{16}$$

Multiplying by $16$, we get $528 = 2t + t^3$, or:

$$t^3 + 2t - 528 = 0$$

### Phase 3: The Algebraic Triumph

I know that seeing a cubic equation can be daunting, but in the world of JEE, there is almost always a hidden elegance. We test integer values. If we try $t = 8$, we calculate:

$$8^3 + 2(8) = 512 + 16 = 528$$

It works perfectly! With $t = 8$, we find the coordinates of $P$:

$$h = \frac{8^2}{16} = 4, \quad k = \frac{8}{8} = 1$$

So, our point $P$ is $(4, 1)$.

### Phase 4: The Second Parabola

Now, we shift our focus to the second curve: $y^2 = 4(x + y)$. To find its directrix, we must reveal its true form. Expanding and rearranging, we get $y^2 - 4y = 4x$.

We complete the square on the left: $y^2 - 4y + 4 = 4x + 4$, which simplifies to:

$$(y - 2)^2 = 4(x + 1)$$

This is a parabola with vertex $(-1, 2)$ and focal length $A = 1$. For a parabola of the form $Y^2 = 4AX$, the directrix is $X = -A$. Here, $X = x + 1$ and $A = 1$, so $x + 1 = -1$, which means the directrix is the vertical line $x = -2$.

### The Final Step

We have our point $P(4, 1)$ and our directrix $x = -2$. The perpendicular distance from a point $(x_1, y_1)$ to a vertical line $x = c$ is simply $|x_1 - c|$.

Thus, the distance is:

$$|4 - (-2)| = |4 + 2| = 6$$

We have arrived at our destination. The beauty of this problem lies in how it forces us to switch between different mathematical tools—normal equations, cubic roots, and coordinate geometry—to find a single, elegant result.

**The final distance is 6.**

Animated Solution for Mathematics - Conic Sections: If $P (h, k)$ be point on the parabola $x = 4 y^{2}$ , which is nearest to the point $Q (0, 33)$ , then the distance of P from the directrix of the parabola $y^{2} = 4 (x + y)$ is equal to :

Select Answer:

Visualized Solution

Analyze Parabola $x = 4 y^{2}$

Shortest Distance to $Q (0, 33)$

Parametric Setup for $P$

Pass Normal through $Q (0, 33)$

Solve the Cubic Equation

Find Coordinates of $P (h, k)$

Analyze Parabola $y^{2} = 4 (x + y)$

Convert to Standard Form

Identify the Directrix

Calculate Final Distance $d$

The Sigma Insight: Equation of Tangent and Normal

The Geometry of Proximity

Phase 1

Phase 2

Phase 3

Phase 4

The Final Step

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